Question

# If n=110 and widehat{p} (p-hat) =0.66, construct a 95% confidence interval.Give your answers to two decimals? <p< ?

Confidence intervals

If $$\displaystyle{n}={110}$$ and $$\displaystyle{w}{i}{d}{e}\hat{{{p}}}$$ (p-hat) $$\displaystyle={0.66}$$, construct a $$\displaystyle{95}\%$$ confidence interval. Give your answers to two decimals $$\displaystyle?{<}{p}{<}?$$

2020-10-26

Step 1 The confidence interval for a sample proportion at $$\displaystyle{95}\%$$ can be calculated using the formula, $$\displaystyle{w}{i}{d}{e}\hat{{{p}}}-{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}{<}{p}{<}{w}{i}{d}{e}\hat{{{p}}}+{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}$$, where $$\displaystyle{w}{i}{d}{e}\hat{{{p}}}$$ is the sample proportion, N is the sample size and z* is the approprite value from the standard normal distributions for the desired confidence intervals. The z* value for $$\displaystyle{95}\%$$ confidence interval is equal to 1.96 Step 2 The sample proportion is equal to 0.66 and the sample size is equal to 110 and the z* value is equal to 1.96 Substitute the values of $$\displaystyle{w}{i}{d}{e}\hat{{{p}}}$$, N and z* in the formula, $$\displaystyle{w}{i}{d}{e}\hat{{{p}}}-{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}{<}{p}{<}{w}{i}{d}{e}\hat{{{p}}}+{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}$$ to calculate the confidence interval of the sample proportion at $$\displaystyle{95}\%$$ interval $$\displaystyle{w}{i}{d}{e}\hat{{{p}}}-{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}{<}{p}{<}{w}{i}{d}{e}\hat{{{p}}}+{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}$$
$$\displaystyle{0.66}-{\left({1.96}\sqrt{{{\frac{{{0.66}{\left({1}-{0.66}\right)}}}{{{110}}}}}}\right)}{<}{p}{<}{0.66}+{\left({1.96}\sqrt{{{\frac{{{0.66}{\left({1}-{0.66}\right)}}}{{{110}}}}}}\right)}$$
$$\displaystyle{0.66}-{\left({1.96}\sqrt{{{0.00204}}}\right)}{<}{p}{<}{0.66}+{\left({1.96}\sqrt{{{0.00204}}}\right)}$$
$$\displaystyle{0.66}-{\left({1.96}{\left({0.0451}\right)}{<}{p}{<}{0.66}+{\left({1.96}{\left({0.0451}\right)}\right.}\right.}$$
$$\displaystyle{0.66}-{0.0884}{<}{p}{<}{0.66}+{0.0884}$$
$$\displaystyle{0.5716}{<}{p}{<}{0.7484}$$ The confidence interval for the sample proportion at $$\displaystyle{95}\%$$ is equal to (0.5716, 0.7484)